What is the height from the surface of earth, where acceleration due to gravity will be $1 / 4$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$

The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5 \mathrm{~ms}^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth= 6400 km )

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ( $R=$ Radius of the earth)

Two satellites of masses $m$ and 1.5 m are revolving around the earth with different speeds in two circular orbits of heights $R_E$ and $2 R_E$ respectively, where $R_F$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces on the earth due to the two satellites is